Experiment 5

Kirchhoff's Rules




The objective of this experiment is to verify Kirchhoff's rules applied to a two-loop circuit.



A few ceramic resistors (200 to 500 ohms), two dc power sources (0 to 20 volts),  5 multi-meters, a calculator, and a few connecting wires with alligator clips




Ohm's law alone is not sufficient to solve for unknown currents in multi-loop circuits.  In such circuits, Kirchhoff's rules are used to solve for the unknown currents.  There are two rules:


Kirchhoff's Loop Rule (KLR)

Kirchhoff's Junction Rule (KJR)


a) KLR:


   This rule states that "the sum of voltage jumps and drops across the elements of any closed loop is zero."

   The voltages in a loop are either 1) due to an existing battery Vbat that we just write its voltage down, or  2) because of the current flowing in a resistor that we write the voltage across that resistor as the product RI.  The product RI is also a Voltage.  In writing a KLR, we just add the battery voltages and the RI voltages for all resistors in the loop and set the total equal to zero.  Of course, we need to be careful which ones are positive and which ones are negative as follows:

1) Rule for Batteries:  As we trace a closed loop in either direction, if we go from (-) to (+) across a battery, it is a voltage jump and we write +Vbat for that battery.  If we happen to go from (+) to (-) of it, it is a voltage drop and we write it as -Vbat . 

2) Rule for Resistors:  If our tracing direction opposes the assumed current I, we write +RI (a voltage jump) across resistor R, and if our tracing direction agrees with I, we write -RI (a voltage drop) across resistor R.

   The reason for the above is that when we swim against the flow of a river, we are trying to gain elevation, potential energy, or simply potential.  You know that water always flows from high elevation to low elevation.  Swimming against a water flow is an effort to gain elevation, or simply potential, and is positive.  Going from one end of a resistor to its other end against current I is like going to a higher potential or a higher voltage point of the loop that means a "voltage jump."  We then write the product RI with a (+) sign.  If our trace direction agrees with the assumed current I, it is like losing potential and the product RI deserves a (-) sign.

   Verify the correctness of Equations (1) and (2) for the 2-loop circuit shown below based on the assumed directions for currents I1 , I2, and I3 .

Figure 1

 KLR in Loop AGBCDA:    (1)

 -V2 +R3I3 +V1 +R2I1 +R1I1 = 0

 KLR in Loop AGBFEA:    (2)

 -V2 + R3I3 - R4I2 +V = 0.     

 KJR at Junction A:          

  - I1  +  I2  +  I3  =  0.   (3)     

Choosing Current Directions:  We don't know the true direction of the currents to begin with We simply assume a direction for the current in each branch.  If our assumption is correct, the answers we get at the end of our calculations will be positive.  If our assumption on a current direction is wrong, the calculations will yield a negative value for that current and we will know it at the end.

b) KJR:

This rule states that "The sum of currents going toward a junction is equal to the sum of currents leaving that junction."   In other words, the algebraic sum of currents to and away from a junction is zero. 

   A junction is a point of connection of 3 or more wires.  Equation (3) above is a KJR written for Junction A in Fig. 1.   I1 leaves Junction A and is given a (-) sign.   Both I2 and I3  go toward Junction A, they are given (+) signs.

A two-loop circuit has three branches.  For each branch, the current must be determined; therefore, there are three unknowns.  Three equations are needed to solve for three unknowns.  Two KLRs and one KJR will provide the three equations.




With R1 , R2 , and R3 chosen between 200 Ω and 500 Ω, construct a two-loop circuit as shown in Fig. 3.   Initially, let V1 = 8V and V2 = 5V.  Change the voltages such that each is different and yet below 15V while you read 3 different currents from the ammeters that none is less than 8mA.

Figure 3


When the circuit is fairly stabilized, write down the values of V1 , V2 , I1 , I2 , and I3 as read from the five multi-meters.  The I1 , I2 , and I3 you read from the meters are your measured values.


Use V1 and V2 in conjunction with R1 , R2 , and R3 as the "given data."   Use these five values to solve for currents I1 , I2 , and I3  by applying Kirchhoff's Rules.  These will be your accepted values for the currents.






R1 =........ Ω ,     R2 =........ Ω     R3 =....... Ω.


    The stabilized V1 = .......... volts  and V2 = ...........volts.




I1 = ......... I2 = ........... I3 = ..........   as read from the ammeters.




With the given values of V1 , V2 , R1 , R2 , and R3, apply KLR and KJR to solve for I1 , I2 , and I3 , and use these calculated values as accepted values.


Comparison of the Results:


Corresponding to every measured value, there is an accepted value.  Calculate a % error on each using the same percent error formula as in Experiment 2.


Conclusion: To be explained by students.


Discussion:  To be explained by students.



Solving System of Equations Using TI 83 and TI 84


To solve the following already rearranged (ordered) system that has 2 unknowns, we need a 2 x 3 matrix,

{ 3x + -2y = 12

6x +  4y  = -3

the A-matrix is:

3 -2

, and the B-matrix is 

6 4 -3


1) TI 83 and 84

To input and solve the matrix, 
1)  Press [2nd] [MATRX] (Note: On the TI-83, just press [MATRX]).
2)  Scroll to Edit.
3)  Press [1] to access matrix A.
4)  Input the dimensions [2] [ENTER] [3] [ENTER].   Note: See (*) below.
5)  Input the matrix entries, pressing enter after each value.
6)  Press [2nd] [QUIT] [2nd] [MATRX] (Press [2nd] [QUIT] [MATRX] on the TI-83).
7)  Scroll to MATH.
8)  Press [ALPHA] [B] [2nd] [MATRX] [1] [)] [ENTER].

     (With the TI-83, press [ALPHA] [B] [MATRX] [1] [)] [ENTER]).

The solution to the above system of 2 equations in 2 unknown should appear as 

x = 1.75 and y = -3.375.

(*) To solve for 2 unknowns, you will need a 2 x 3 matrix).

      To solve for 3 unknowns, you will need a 3 x 4 matrix).

      To solve for 4 unknowns, you will need a 4 x 5 matrix) and so on.


Solving System of Equations Using TI 85 and 86


In the already ordered system:

3x + -2y = 12
6x +  4y  = -3 ,  the 1st row elements are: 3    -2  & 12

                    and the 2nd row elements are: 6     4  & -3.


1) Press [2nd] [TABLE]  to pop up the SIMULT for simultaneous equations.

2)  Enter the # of equs. to be solved.  For this example:  Number = 2  then Press [ENTER].

3) You will be asked for elements of the 1st row, here:  3   -2  &  12  Press [ENTER] after each entry.

4) You will be asked for elements of the 2nd row, here:  6   4  &  -3  Press [ENTER] after each entry.

5) Press [F5] to Solve.  You should get: x = 1.75 and y = -3.375.